Find the sum of the cubes of first n even number, and first n odd number.


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Question Number 57635 by Tawa1 last updated on 09/Apr/19

$Find the sum of the cubes of first n even number, and$ $first n odd number .$
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
$As per higher algebra Bernard and child we can find s_r =1^r +2^r +3^r +...+n^r by intregation snd intregation constant as Bernuli numbers example s_1 =1+2+3+...+n =((n(n+1))/2)=(n^2 /2)+(n/2) now to find S_2 =1^2 +2^2 +3^2 +...+n^2 using S_1 S_r =r∫S_(r−1) dn+nB_r ←formula S_2 =2∫((n^2 /2)+(n/2))+n×(1/6) [B_2 =(1/6)] =2((n^3 /6)+(n^2 /4))+n×(1/6) =(n^3 /3)+(n^2 /2)+(n/6) =((2n^3 +3n^2 +n)/6) =(n/6)×(2n^2 +2n+n+1) =(n/6)×{2n(n+1)+1(n+1)} =((n(n+1)(2n+1))/6) similarly S_3 =3∫S_2 dn+nB_3 =3∫((n^3 /3)+(n^2 /2)+(n/6))dn+n×0 =(n^4 /4)+(n^3 /2)+(n^2 /4) =((n^4 +2n^3 +n^2 )/4) =((n^2 (n^2 +2n+1))/4) ={((n(n+1))/2)}^2 thus you can find...$
$A s p e r h i g h e r a l g e b r a B e r n a r d a n d c h i l d$ $w e c a n f i n d$ $s_{r} = 1^{r} + 2^{r} + 3^{r} + . . . + n^{r}$ $b y i n t r e g a t i o n s n d i n t r e g a t i o n c o n s t a n t$ $a s B e r n u l i n u m b e r s$ $e x a m p l e$ $s_{1} = 1 + 2 + 3 + . . . + n$ $= \frac{n (n + 1)}{2} = \frac{n^{2}}{2} + \frac{n}{2}$ $n o w t o f i n d S_{2} = 1^{2} + 2^{2} + 3^{2} + . . . + n^{2}$ $u s i n g S_{1}$ $S_{r} = r \int S_{r - 1} d n + {n B}_{r} \leftarrow f o r m u l a$ $S_{2} = 2 \int (\frac{n^{2}}{2} + \frac{n}{2}) + n \times \frac{1}{6} [B_{2} = \frac{1}{6}]$ $= 2 (\frac{n^{3}}{6} + \frac{n^{2}}{4}) + n \times \frac{1}{6}$ $= \frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}$ $= \frac{2 n^{3} + 3 n^{2} + n}{6}$ $= \frac{n}{6} \times (2 n^{2} + 2 n + n + 1)$ $= \frac{n}{6} \times {2 n (n + 1) + 1 (n + 1)}$ $= \frac{n (n + 1) (2 n + 1)}{6}$ $s i m i l a r l y$ $S_{3} = 3 \int S_{2} d n + {n B}_{3}$ $= 3 \int (\frac{n^{3}}{3} + \frac{n^{2}}{2} + \frac{n}{6}) d n + n \times 0$ $= \frac{n^{4}}{4} + \frac{n^{3}}{2} + \frac{n^{2}}{4}$ $= \frac{n^{4} + 2 n^{3} + n^{2}}{4}$ $= \frac{n^{2} (n^{2} + 2 n + 1)}{4}$ $= {\frac{n (n + 1)}{2}}^{2}$ $t h u s y o u c a n f i n d . . .$
Commented by Tawa1 last updated on 09/Apr/19

$God bless you sir . I appreciate .$
	Answered by mr W last updated on 09/Apr/19

	$S_{E} = \underset{k = 1}{\sum^{n}} {(2 k)}^{3} = 8 \underset{k = 1}{\sum^{n}} k^{3} = 8 \times \frac{n^{2} {(n + 1)}^{2}}{4}$ $= 2 n^{2} {(n + 1)}^{2}$ $S_{O} = \underset{k = 1}{\sum^{n}} {(2 k - 1)}^{3} = \underset{k = 1}{\sum^{n}} (8 k^{3} - 12 k^{2} + 6 k - 1) = 8 \times \frac{n^{2} {(n + 1)}^{2}}{4} - 12 \times \frac{n (n + 1) (2 n + 1)}{6} + 6 \times \frac{n (n + 1)}{2} - n$ $= n^{2} (2 n^{2} - 1)$
	Commented by Tawa1 last updated on 09/Apr/19

	$Wow, God bless you sir . I appreciate .$
	Commented by Tawa1 last updated on 09/Apr/19

	$But sir, i have a question .$ $Please see this question .$ $simplify : \frac{\sqrt{10 + \sqrt{1}} + \sqrt{10 + \sqrt{2}} + . . . + \sqrt{10 + \sqrt{99}}}{\sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + . . . + \sqrt{10 - \sqrt{99}}}$ $Answer : \sqrt{2} + 1$
	Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

	$e x c e l l e n t s i r . . .$
	Commented by Tawa1 last updated on 09/Apr/19

	$God bless you sir . But i have checked . i usually undersand your$ $solution better sir .$
	Commented by mr W last updated on 09/Apr/19

	$i l i k e t h i s s o l u t i o n :$ $A = \underset{k = 1}{\sum^{99}} \sqrt{10 + \sqrt{k}}$ $B = \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}}$ ${(\sqrt{10 + \sqrt{k}} - \sqrt{10 - \sqrt{k}})}^{2} = 10 + \sqrt{k} - 2 \sqrt{(10 + \sqrt{k}) (10 - \sqrt{k})} + 10 - \sqrt{k}$ ${(\sqrt{10 + \sqrt{k}} - \sqrt{10 - \sqrt{k}})}^{2} = 2 (10 - \sqrt{100 - k})$ $\Rightarrow \sqrt{10 + \sqrt{k}} - \sqrt{10 - \sqrt{k}} = \sqrt{2} \sqrt{10 - \sqrt{100 - k}}$ $\Rightarrow \underset{k = 1}{\sum^{99}} \sqrt{10 + \sqrt{k}} - \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}} = \sqrt{2} \underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}}$ $\Rightarrow A - B = \sqrt{2} B$ $\Rightarrow A = (\sqrt{2} + 1) B$ $\Rightarrow \frac{A}{B} = \sqrt{2} + 1$
	Commented by Tawa1 last updated on 09/Apr/19

	$God bless you sir . I appreciate .$
	Commented by Tawa1 last updated on 09/Apr/19

	$Sir, you said B = \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}}$ $finally how is \underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}} = B again . . .$
	Commented by mr W last updated on 09/Apr/19

	$g e n e r a l l y :$ $\frac{\underset{k = m}{\sum^{n^{2} - m}} \sqrt{n + \sqrt{k}}}{\underset{k = m}{\sum^{n^{2} - m}} \sqrt{n - \sqrt{k}}} = \sqrt{2} + 1 w i t h n, m \in N$
	Commented by Tawa1 last updated on 09/Apr/19

	$wow . I understand now . God bless you .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

	$l o o k . . . B (w h e n k \to 1 \to 99) = \underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}}$ $B (k \to 1 \to 99 = \sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + \sqrt{10 - \sqrt{3}} + . . + \sqrt{10 - \sqrt{99}}$ $n o w l o o k$ $B (k \to 99 \to 1) [\underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}}]$ $= \sqrt{10 - \sqrt{1}} + \sqrt{10 - \sqrt{2}} + . . . + \sqrt{10 - \sqrt{99}}$ $h e n c e$ $\underset{k = 1}{\sum^{99}} \sqrt{10 - \sqrt{k}} = \underset{k = 99}{\sum^{1}} \sqrt{10 - \sqrt{100 - k}}$
	Commented by Meritguide1234 last updated on 10/Apr/19

	$see q . 54226 (ans already provided)$
	Commented by malwaan last updated on 10/Apr/19

	$thank you$
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